If an isosceles triangle ABC , in which AB = AC = 10 cm , is inscribed in a circle of radius 10 cm , find the area of the triangle .

Answer

To find the area of the isosceles triangle ABC inscribed in a circle of radius 10 cm, where AB = AC = 10 cm, we can follow these steps:

### Step 1: Understand the Geometry
We have an isosceles triangle ABC with AB = AC = 10 cm. The circle in which the triangle is inscribed has a radius of 10 cm. The center of the circle is O.

### Step 2: Identify Key Points
Let D be the midpoint of BC. Since triangle ABC is isosceles, AD (the height from A to BC) will also bisect BC. Thus, BD = DC = x.

### Step 3: Use the Pythagorean Theorem
In triangle ADB, we can apply the Pythagorean theorem:
\[ AB^2 = AD^2 + BD^2 \]
Substituting the known values:
\[ 10^2 = AD^2 + x^2 \]
This simplifies to:
\[ 100 = AD^2 + x^2 \] (Equation 1)

### Step 4: Use the Circle's Radius
In triangle ODB, where O is the center of the circle, we also apply the Pythagorean theorem:
\[ OB^2 = OD^2 + BD^2 \]
Since OB is the radius of the circle:
\[ 10^2 = OD^2 + x^2 \]
This simplifies to:
\[ 100 = OD^2 + x^2 \] (Equation 2)

### Step 5: Express OD in Terms of AD
Since D is the midpoint of BC, we can express OD in terms of AD:
\[ OD = 10 - AD \]
Substituting this into Equation 2 gives:
\[ 100 = (10 - AD)^2 + x^2 \]
Expanding this:
\[ 100 = 100 - 20AD + AD^2 + x^2 \]
This simplifies to:
\[ 20AD = AD^2 + x^2 \] (Equation 3)

### Step 6: Solve the Equations
Now we have two equations:
1. \( 100 = AD^2 + x^2 \) (Equation 1)
2. \( 20AD = AD^2 + x^2 \) (Equation 3)

From Equation 1, we can express \( x^2 \) as:
\[ x^2 = 100 - AD^2 \]
Substituting this into Equation 3:
\[ 20AD = AD^2 + (100 - AD^2) \]
This simplifies to:
\[ 20AD = 100 \]
Thus:
\[ AD = 5 \]

### Step 7: Find x
Substituting \( AD = 5 \) back into Equation 1 to find \( x \):
\[ 100 = 5^2 + x^2 \]
\[ 100 = 25 + x^2 \]
\[ x^2 = 75 \]
Thus:
\[ x = 5\sqrt{3} \]

### Step 8: Calculate the Area of Triangle ADB
The area of triangle ADB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, base \( BD = x = 5\sqrt{3} \) and height \( AD = 5 \):
\[ \text{Area}_{ADB} = \frac{1}{2} \times 5\sqrt{3} \times 5 = \frac{25\sqrt{3}}{2} \]

### Step 9: Find the Area of Triangle ABC
Since triangle ABC is composed of two triangles ADB and ADC:
\[ \text{Area}_{ABC} = 2 \times \text{Area}_{ADB} = 2 \times \frac{25\sqrt{3}}{2} = 25\sqrt{3} \]

### Final Answer
The area of triangle ABC is:
\[ \text{Area}_{ABC} = 25\sqrt{3} \, \text{cm}^2 \]